MHB Justine's question at Yahoo Answers regarding the Midpoint and Trapezoidal Rules

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The discussion focuses on demonstrating the relationship between the Midpoint and Trapezoidal Rules in numerical integration. It begins with the definitions of both rules, outlining how each approximates the integral of a function over an interval. By dividing the interval into 2n partitions, the Midpoint Rule is reformulated to align with the Trapezoidal Rule's structure. The key finding is that combining the two estimates leads to the conclusion that half the sum of the Trapezoidal and Midpoint estimates equals the Trapezoidal estimate with double the partitions, expressed as (1/2)(Tn + Mn) = T2n. This relationship highlights the connection between the two numerical methods for approximating integrals.
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Here is the question:

Trapezoidal and Midpoint Estimates?


Show that (1/2)(Tn+Mn)=T2n

I have posted a link there to this topic so the OP can see my work.
 
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Hello Justine,

First, let's look at the definitions of the two rules:

Midpoint Rule:

The Midpoint Rule is the approximation $$\int_a^b f(x)\,dx\approx M_n$$, where

$$M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_1}{2} \right)+f\left(\frac{x_1+x_2}{2} \right)+\cdots+f\left(\frac{x_{n-1}+x_{n}}{2} \right) \right]$$

Trapezoidal Rule:

The Trapezoidal Rule is the approximation $$\int_a^b f(x)\,dx\approx T_n$$, where

$$T_n=\frac{b-a}{2n}\left[f\left(x_0 \right)+2f\left(x_1 \right)+\cdots+2f\left(x_{n-1} \right)+f\left(x_{n} \right) \right]$$

We can divide the interval $[a,b]$ into $2n$ equally spaced partitions, as so we may then write:

$$M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_2}{2} \right)+f\left(\frac{x_2+x_4}{2} \right)+\cdots+f\left(\frac{x_{2n-2}+x_{2n}}{2} \right) \right]$$

And so using the midpoint formula, this becomes:

$$M_n=\frac{b-a}{2n}\left[2f\left(x_1 \right)+2f\left(x_3 \right)+\cdots+2f\left(x_{2n-1} \right) \right]$$

Likewise, the Trapezoidal Rule may now be written:

$$T_n=\frac{b-a}{2n}\left[f\left(x_0 \right)+2f\left(x_2 \right)+\cdots+2f\left(x_{2n-2} \right)+f\left(x_{2n} \right) \right]$$

Multiplying both by $$\frac{1}{2}$$ and adding, we find:

$$\frac{1}{2}\left(T_n+M_n \right)=\frac{b-a}{2(2n)}\left[f\left(x_0 \right)+2f\left(x_1 \right)+\cdots+2f\left(x_{2n-1} \right)+f\left(x_{2n} \right) \right]$$

And using the definition of the Trapezoidal Rule, we find:

$$\frac{1}{2}\left(T_n+M_n \right)=T_{2n}$$
 
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